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Let $V$ and $W$ be Brownian motions such that $\mathbb{E}W_tV_t=\rho t$. Let $$R_t=\sup_{u \le t} V_u \mbox{ and } Z_t=\sup_{u \le t}W_u .$$

Show that $$\mathbb{E}R_tZ_t=tf(\rho) .$$

Can you find $f(\rho)$ ?

EDIT:

I have already figured out the first part of the question. Let us fix $t > 0$ and define $\overline{W}_s=\sqrt{t} \cdot W_{\frac{s}{t}}$ and $\overline{V}_s=\sqrt{t} \cdot V_{\frac{s}{t}}$. Let us notice that $\overline{W},\overline{V}$ are both Brownian motions and $\mathbb{E}\overline{W}_s\overline{V}_s=\rho s$. Therefore $\left( W,V \right)\stackrel{d}{=}\left( \overline{W},\overline{V} \right)$.

Hence for $\overline{R}_t=\sup_{u \le t} \overline{V}_u \mbox{ and } \overline{Z}_t=\sup_{u \le t}\overline{W}_u$ we have $\left( R,Z \right)\stackrel{d}{=}\left( \overline{R},\overline{Z} \right)$ and $\overline{R}_t=\sup_{u \le t}\overline{V}_u=\sup_{u \le t} \sqrt{t} \cdot V_{\frac{u}{t}}=\sqrt{t} \cdot R_1$ and analogously $\overline{Z}_t=\sqrt{t} \cdot Z_1$.

Finally $$\mathbb{E}R_tZ_t=\mathbb{E}\overline{R}_t\overline{Z}_t=\mathbb{E} \sqrt{t} \cdot R_1 \cdot \sqrt{t} \cdot Z_1=t\mathbb{E}R_1Z_1$$

rafalpw
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  • What are your thoughts on the problem? – Rory Daulton Feb 12 '16 at 15:31
  • I figured out the first part of the question, but I can't post the answer, because the question is now on hold – rafalpw Feb 14 '16 at 12:30
  • You can and should edit your question to add that information to your question. Just click the "edit" link immediately below your question. If your edit is good enough you will get the votes needed to reopen your question. – Rory Daulton Feb 14 '16 at 12:32
  • But I don't think there's anything missing in the question. The problem seems quite clear to me. – rafalpw Feb 14 '16 at 12:34
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    As the hold notice states, "...which ideally includes your thoughts on the problem and any attempts you have made to solve it." It is not enough that the question is clear: we also want to see that you have done significant work on the problem. This is not a homework-answering site, so we want to filter out lazy questions. Show us that you have done sufficient work, then we will be happy to help you. – Rory Daulton Feb 14 '16 at 12:38

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